4.6 Electromagnetic induction
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
Unit 4 (WPH14), A2 content: specification points 108–110. Flux density and the force on a current or a moving charge are on 4.5 Magnetic fields and forces; peak and r.m.s. values of alternating current are not part of this topic.
What this page covers
8 points
- A short recap of magnetic flux and flux linkage, the quantities this topic is built on
- The factors that decide the e.m.f. induced when a magnet and a coil move relative to each other
- The e.m.f. induced in a coil by a changing current in a linked coil, and how a transformer works (qualitatively)
- Faraday's law and Lenz's law: \(\varepsilon = -\dfrac{d(N\phi)}{dt}\)
- Flux-linkage–time and e.m.f.–time graphs: the e.m.f. is minus the gradient
- A straight conductor moving through a uniform field: \(\varepsilon = Blv\)
- A coil rotating in a uniform field and its sinusoidal e.m.f. (qualitatively)
- Eddy currents and magnetic braking as applications of Lenz's law
In one minute
6 points
- An e.m.f. is induced in a circuit whenever the magnetic flux linking it changes, and its size equals the rate of change of flux linkage while its direction opposes the change: \(\varepsilon = -d(N\phi)/dt\).
- Flux \(\phi = BA\) (field perpendicular to the area) is in webers, Wb = T m²; flux linkage is \(N\phi\) for a coil of \(N\) turns.
- The flux linkage can change by moving a magnet or the coil, changing a current nearby, or rotating the coil; a steady flux linkage, however large, induces nothing.
- On a graph of \(N\phi\) against \(t\), the e.m.f. is minus the gradient; the area under an \(\varepsilon\)–\(t\) graph equals the change in flux linkage in size.
- A rod of length \(l\) moving at \(v\) at right angles to a field \(B\) has \(\varepsilon = Blv\); a coil turning steadily gives a sinusoidal e.m.f., largest when the coil is edge-on to the field.
- Eddy currents in solid metal oppose relative motion (magnetic braking) and waste energy as heat in transformer cores.
Contents
14 sections
- What are magnetic flux and flux linkage?
- What happens when a magnet moves relative to a coil?
- How can a changing current in one coil induce an e.m.f. in another?
- What do Faraday's law and Lenz's law say?
- How do you use flux-linkage–time and e.m.f.–time graphs?
- What e.m.f. is induced in a rod moving through a field?
- Why does a rotating coil give an alternating e.m.f.?
- What are eddy currents and how do they brake motion?
- Exam tips for 4.6
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What are magnetic flux and flux linkage?
These are defined on 4.5 Magnetic fields and forces; here is what this topic needs from them.
Think of flux as the amount of field passing through an area. If the area is tilted, less field passes through it; when the plane of the coil is parallel to the field lines, the flux through it is zero. Each turn of a coil has the same flux through it, so a coil of \(N\) turns has \(N\) times the flux linkage of a single turn. Flux linkage can change in three ways: change \(B\), change the area inside the field, or change the angle between the coil and the field. Every example on this page is one of these.
What happens when a magnet moves relative to a coil?
Connect a coil to a sensitive centre-zero meter and push a bar magnet into it: the needle flicks one way. Hold the magnet still: the reading returns to zero. Pull it out: the needle flicks the other way. Moving the coil over a stationary magnet does exactly the same, so only the relative motion matters.
The e.m.f. exists whenever the flux linkage is changing, even if the coil is not part of a complete circuit. A current flows only if the circuit is complete.
| Change made | Effect on the induced e.m.f. | Reason |
|---|---|---|
| Move the magnet faster | larger | the flux linkage changes in less time |
| Use a stronger magnet | larger | bigger change in flux for the same movement |
| Use a coil with more turns | larger | each turn adds its own e.m.f.: flux linkage is \(N\phi\) |
| Use a coil of larger cross-sectional area | larger (for a wide enough field) | more flux passes through each turn |
| Reverse the direction of motion, or turn the magnet round | reverses | the flux linkage now changes the other way |
| Hold the magnet still inside the coil | zero | flux linkage is large but not changing |
Saying that a magnet resting inside a coil induces an e.m.f. because "there is a lot of flux". The e.m.f. depends on the rate of change of flux linkage, not on its size. A large, steady flux linkage induces nothing.
How can a changing current in one coil induce an e.m.f. in another?
A current in a primary coil makes a magnetic field. If a secondary coil sits in that field, best of all wound on the same iron core, some of the primary's flux links the secondary. When the primary current changes, the flux linking the secondary changes, so an e.m.f. is induced in the secondary. Nothing needs to move. A steady primary current gives a steady flux and no e.m.f. in the secondary.
What makes the secondary e.m.f. larger?
- a faster rate of change of the primary current (switching more quickly, or a higher a.c. frequency)
- a larger change in the primary current
- more turns on the primary (stronger field for the same current) or on the secondary (more flux linkage)
- an iron core linking the two coils: iron concentrates the flux and guides almost all of it through the secondary
- placing the coils closer together, so more of the primary's flux passes through the secondary
Transformers (qualitatively)
An alternating p.d. across the primary drives an alternating current, which produces a continuously changing flux in the iron core. The core carries this changing flux through the secondary, so an alternating e.m.f. of the same frequency is induced across it. More secondary turns than primary turns gives a larger secondary e.m.f. (step-up); fewer gives a smaller one (step-down). A transformer will not work on a steady d.c. supply: the flux is constant, so no e.m.f. is induced in the secondary, apart from a pulse at switch-on and switch-off.
Two coils are wound on the same iron core, as in the figure above. The primary is connected to a cell and a switch; the secondary is connected to a voltmeter. Describe and explain what the voltmeter shows (a) as the switch is closed, (b) while it stays closed and (c) as it is opened.
(a) The primary current grows from zero, so the flux in the core increases. The changing flux linkage of the secondary induces an e.m.f.: a brief reading
(b) The current is steady, so the flux is constant and the flux linkage is not changing: reading zero
(c) The current falls to zero, so the flux decreases, which is the opposite change, and a faster one: a brief reading in the opposite direction, larger than in (a), because the flux changes more quickly.
What do Faraday's law and Lenz's law say?
The experiments above all point to two rules: one for the size of the e.m.f. and one for its direction.
Lenz's law: the direction of the induced e.m.f. is such that any current it drives opposes the change that produced it.
Why Lenz's law must be true
Suppose the induced current helped the change. Pushing a magnet in would make the coil attract it, so the magnet would speed up, inducing more current, attracting it more, and so on: electrical energy and kinetic energy from nothing. Instead the induced current always makes you do work against a magnetic force, and that work is the source of the electrical energy. Lenz's law is conservation of energy.
Using it: push a north pole into a coil and the near end of the coil becomes a north pole, which repels it. Pull the north pole away and the near end becomes a south pole, which attracts it and so opposes the withdrawal.
The north pole of a magnet is pushed into a coil of 250 turns and cross-sectional area 3.2 × 10⁻⁴ m². The flux density inside the coil rises from zero to 0.060 T in 0.15 s. (a) Calculate the average e.m.f. induced. (b) State the polarity of the end of the coil nearest the magnet. (c) The magnet is then pulled out in 0.050 s. Calculate the new average e.m.f.
Change in flux linkage: \(\Delta(N\phi) = N \times \Delta B \times A = 250 \times 0.060 \times 3.2\times10^{-4} = 4.8\times10^{-3}\) Wb.
(a) \(\varepsilon = 4.8\times10^{-3} \div 0.15\) = 0.032 V (32 mV)
(b) By Lenz's law the coil opposes the approaching north pole: the near end is a north pole
(c) Same change in flux linkage in one-third of the time: \(\varepsilon = 4.8\times10^{-3} \div 0.050\) = 0.096 V, in the opposite direction.
Flux is a property of one turn; Faraday's law uses the flux linkage, \(N\phi\). Leaving out the 250 here gives 1.3 × 10⁻⁴ V, which is 250 times too small.
How do you use flux-linkage–time and e.m.f.–time graphs?
| Graph | Gradient gives | Area under the line gives |
|---|---|---|
| flux linkage \(N\phi\) against time | minus the e.m.f. (\(\varepsilon\) = −gradient) | no useful quantity |
| e.m.f. \(\varepsilon\) against time | no useful quantity at this level | minus the change in flux linkage |
Where \(N\phi\) is constant (a horizontal line, or a maximum or minimum of a curve) the e.m.f. is zero. Where the \(N\phi\) graph is steepest, the e.m.f. is largest. A rising \(N\phi\) gives one sign of e.m.f. and a falling \(N\phi\) the other. For a curve, find the e.m.f. at an instant from the gradient of the tangent.
The flux-linkage graph above shows how the flux linkage of a coil varies with time. (a) Determine the e.m.f. induced between 0 and 0.20 s, and between 0.50 s and 0.60 s. (b) The coil is part of a circuit of total resistance 2.5 Ω. Calculate the current between 0.50 s and 0.60 s.
(a) \(\varepsilon\) = −gradient. For 0–0.20 s: \(\varepsilon = -(0.030 - 0) \div (0.20 - 0)\) = −0.15 V. For 0.50–0.60 s: \(\varepsilon = -(0 - 0.030) \div (0.60 - 0.50)\) = +0.30 V, twice as large and of opposite sign.
(b) \(I = \varepsilon/R = 0.30 \div 2.5\) = 0.12 A
A magnet falling through a coil
As the magnet enters, the flux linkage rises, so the e.m.f. has one sign; as it leaves, the flux linkage falls and the e.m.f. reverses. At the instant the magnet is at the centre of the coil, the flux linkage is at its maximum and momentarily not changing, so the e.m.f. is zero.
The magnet is accelerating, so the flux linkage falls faster than it rose: the second peak is larger and lasts a shorter time. The area under each pulse equals the change in flux linkage during that pulse. The flux linkage starts and ends close to zero, so the two areas are very nearly equal in size.
When you sketch an e.m.f.–time graph from a flux-linkage graph, line the two up vertically. Put a zero of e.m.f. under every peak, trough and flat section of \(N\phi\), and the largest e.m.f. under every steepest part.
What e.m.f. is induced in a rod moving through a field?
A straight conductor moving across a field "cuts" flux. The free electrons inside it move with it, so each feels a magnetic force along the rod: charge separates and an e.m.f. appears between the ends. The same result comes from Faraday's law. In a time \(\Delta t\) a rod of length \(l\) moving at speed \(v\) sweeps out an area \(lv\Delta t\), so the flux it cuts is \(Blv\Delta t\).
A metal rod rests across two horizontal rails 0.25 m apart, joined at one end by a resistor. The total resistance of the circuit is 0.50 Ω. A uniform vertical field of 0.40 T acts over the rails. The rod is pulled along the rails at a steady 3.0 m s⁻¹. Calculate (a) the e.m.f., (b) the current, (c) the force needed to keep the rod moving at constant speed (ignore friction) and (d) the power supplied by that force. Comment on your answer to (d).
(a) \(B\), \(l\) and \(v\) are mutually perpendicular: \(\varepsilon = Blv = 0.40 \times 0.25 \times 3.0\) = 0.30 V
(b) \(I = 0.30 \div 0.50\) = 0.60 A
(c) The current in the rod feels a force \(BIl\) opposing the motion (Lenz); at constant speed the pull must balance it: \(F = BIl = 0.40 \times 0.60 \times 0.25\) = 0.060 N
(d) \(P = Fv = 0.060 \times 3.0\) = 0.18 W. This equals \(I^2R = 0.60^2 \times 0.50 = 0.18\) W: every joule of work done pulling the rod is dissipated in the resistance. Lenz's law is energy conservation.
Use Lenz's law in two steps and write both: (1) say what is changing ("the flux linkage into the page is increasing"), (2) say what the induced current must do ("so it flows to create flux out of the page: anticlockwise"). The force on the rod then always opposes its motion.
Why does a rotating coil give an alternating e.m.f.?
A flat coil turning at a steady rate in a uniform field has a flux linkage that varies like a cosine: maximum when the coil faces the field, zero when it is edge-on, and maximum the other way round half a turn later. The e.m.f. is minus the gradient of this curve, so it varies like a sine: an alternating e.m.f. with the same frequency as the rotation. This is the principle of the a.c. generator (alternator).
The e.m.f. is zero when the coil is face-on to the field (flux linkage at a maximum, momentarily not changing) and largest when the coil is edge-on (flux linkage zero, but changing fastest, because the sides of the coil are cutting straight across the field lines).
| Change | Peak e.m.f. | Frequency |
|---|---|---|
| Turn the coil faster | larger (flux linkage changes faster) | larger |
| More turns, larger area or stronger field | larger (larger maximum flux linkage) | unchanged |
For the coil in the figures above (200 turns, area 2.5 × 10⁻³ m², field 0.080 T, 20 revolutions per second): (a) calculate the maximum flux linkage; (b) calculate the average e.m.f. while the coil turns from face-on to edge-on; (c) explain why the peak e.m.f. on the graph is larger than this average; (d) state the effect on the graph of doubling the rate of rotation.
(a) Face-on, the whole area is at right angles to the field: \(N\phi = NBA = 200 \times 0.080 \times 2.5\times10^{-3}\) = 0.040 Wb
(b) \(T = 1 \div 20 = 0.050\) s, so a quarter turn takes 0.0125 s while the flux linkage falls from 0.040 Wb to zero: \(\varepsilon = 0.040 \div 0.0125\) = 3.2 V
(c) The rate of change is not constant: the flux linkage changes slowly near face-on and fastest at edge-on, so the instantaneous e.m.f. there (about 5.0 V) exceeds the average.
(d) The same change in flux linkage happens in half the time: the peak e.m.f. doubles and the period halves (the frequency doubles).
Saying the e.m.f. is largest when the coil is face-on "because the flux is biggest". At that instant the flux linkage is at a turning point, so its gradient, and the e.m.f., is zero. A large e.m.f. needs a large rate of change.
What are eddy currents and how do they brake motion?
Any piece of metal in a changing flux has e.m.f.s induced in it. In a solid conductor these drive loops of current inside the metal itself: eddy currents. By Lenz's law, eddy currents flow so as to oppose the change. If the change comes from motion, the result is a force that opposes the relative motion, and the kinetic energy removed becomes thermal energy in the metal.
Magnetic braking
- Train and ride brakes: strong magnets near a moving metal fin or rail induce eddy currents that slow it without any contact or wear. The braking force grows with speed, so it is gentle at low speeds and cannot bring the object fully to rest on its own.
- Damping a swinging plate: a solid aluminium plate swinging between magnet poles stops within a few swings; a plate with slots cut in it swings for much longer, because the slots break up the current loops.
- A magnet falling down a metal tube falls far more slowly than down a plastic tube (below).
Eddy currents as a nuisance
The alternating flux in a transformer core induces eddy currents in the iron, which waste energy as heat. Cores are made from thin laminated sheets, insulated from each other: the insulation breaks the large current loops, so the eddy currents and the energy wasted are much smaller.
A small strong magnet is dropped down a vertical copper tube 1.20 m long. It takes about 2.5 s to emerge. The same magnet dropped down a plastic tube of the same length takes 0.49 s. Explain these observations.
1. As the magnet falls, the flux linking each ring of the tube changes, so an e.m.f. is induced in it (Faraday).
2. Copper is a good conductor, so these e.m.f.s drive large eddy currents around the tube.
3. The currents oppose the change (Lenz): the ring below repels the approaching magnet and the ring above attracts it, giving an upward force on the magnet.
4. The faster the magnet falls, the faster the flux changes, so the larger the eddy currents and the upward force. It soon equals the weight, the resultant force is zero and the magnet falls at a constant (terminal) velocity.
5. Gravitational potential energy is transferred to thermal energy in the tube instead of to kinetic energy of the magnet.
6. Plastic is an insulator, so no eddy currents flow and there is no braking force: the magnet is in free fall, \(t = \sqrt{2 \times 1.20 \div 9.81}\) = 0.49 s
Long explanations on induction follow the same chain: change of flux linkage → e.m.f. (Faraday) → current, if there is a complete path → direction opposes the change (Lenz) → force or energy consequence. Name each link rather than jumping from "the magnet moves" straight to "it slows down".
Exam tips for 4.6
Every explanation of an induced e.m.f. must contain the idea of a changing flux linkage (or of a conductor cutting flux). "There is a magnetic field" is not enough.
An e.m.f. is induced whenever the flux linkage changes. A current flows only if there is a complete circuit. Read the question to see which one it asks about.
Write \(N\phi = NBA\) as a separate line before dividing by time, and check that \(A\) is in m²: 1 cm² = 1 × 10⁻⁴ m².
\(\varepsilon\) = −gradient of \(N\phi\)–\(t\). A sine-shaped flux linkage gives a cosine-shaped e.m.f. and vice versa; a straight-line section gives a constant e.m.f.; a flat section gives zero.
When asked for a direction or a pole, state what change is happening and that the induced current opposes it. A bare "north pole" or "clockwise" without the reason is an incomplete answer.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A large flux linkage means a large e.m.f. | Faraday's law involves the rate of change, not the size. | A steady flux linkage, however large, induces no e.m.f. |
| A transformer works with a steady d.c. supply. | A steady current gives a constant flux in the core. | Only a changing (e.g. alternating) primary current induces a secondary e.m.f. |
| Lenz's law means the induced current opposes the magnet's field. | It opposes the change in flux, not the field itself. | When the flux is decreasing, the induced field acts in the same direction as the magnet's field. |
| A rotating coil gives its largest e.m.f. when face-on to the field. | There the flux linkage is at a maximum and not changing. | The e.m.f. is largest when the coil is edge-on (flux linkage zero, changing fastest). |
| An e.m.f. needs a complete circuit. | The e.m.f. comes from the changing flux linkage alone. | A complete circuit is needed only for a current. |
| Magnetic braking works because the magnet attracts the metal. | Copper and aluminium are not attracted to magnets. | The braking force comes from induced eddy currents opposing the relative motion. |
How this topic connects
4.5 Magnetic fields and forces
The force \(F = BIl\) on the induced current is what opposes the motion in every Lenz's-law example, and \(\phi = BA\) is defined there.
Electric circuits (Unit 2) and energy (Unit 1)
An induced e.m.f. drives a current through the circuit resistance exactly like a cell; the power dissipated is \(I^2R\). Lenz's law is conservation of energy, and the magnet in a copper tube reaches a terminal velocity for the same reason a ball does in oil: a resistive force that grows with speed.
Oscillations (Unit 5)
Eddy-current damping is a practical way of removing energy from an oscillating system, and a rotating coil's output varies sinusoidally like simple harmonic motion.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\varepsilon = -d(N\phi)/dt\) | Faraday's and Lenz's laws: induced e.m.f. (V) | Yes |
| \(\varepsilon = -\Delta(N\phi)/\Delta t\) | average e.m.f. over a time interval | No — derive it from the line above |
| \(\phi = BA\), flux linkage = \(N\phi\) | flux (Wb) through a coil at right angles to the field | Yes |
| \(\varepsilon = Blv\) | straight conductor moving across a field | No — derive it from Faraday's law |
| \(F = BIl\) | force on the induced current in a conductor (4.5) | Yes |
| \(P = Fv = I^2R\) | energy check for a conductor pulled at constant speed | No — combine \(P = W/t\) and \(P = I^2R\) |
Check your understanding
1. A magnet is held at rest inside a coil connected to a meter. What does the meter read, and why?
Zero. The flux linkage is large but not changing, and the e.m.f. depends on the rate of change of flux linkage.
2. A metal rod 0.30 m long moves at 6.0 m s⁻¹ at right angles to a 0.050 T field. Calculate the e.m.f. between its ends.
\(\varepsilon = Blv = 0.050 \times 0.30 \times 6.0\) = 0.090 V.
3. A coil of 500 turns and area 2.0 × 10⁻⁴ m² is in a field that falls steadily from 0.24 T to zero in 20 ms. Calculate the e.m.f.
\(\Delta(N\phi) = 500 \times 0.24 \times 2.0\times10^{-4} = 0.024\) Wb; \(\varepsilon = 0.024 \div 0.020\) = 1.2 V.
4. Why is a transformer core made of thin insulated sheets rather than solid iron?
The changing flux induces eddy currents in the core. The insulated laminations break up the current loops, so the eddy currents, and the energy wasted heating the core, are much smaller.
5. The flux linkage of a coil falls steadily from 0.012 Wb to zero in 40 ms. Calculate the e.m.f. and sketch the e.m.f.–time graph for this interval.
\(\varepsilon\) = −gradient = \(-(0 - 0.012) \div 0.040\) = +0.30 V: a horizontal line at 0.30 V for the 40 ms, dropping to zero when the flux linkage stops changing.
6. A north pole is pulled away from the end of a coil. State and explain the polarity of that end of the coil while the magnet moves.
The flux linkage is decreasing, so the induced current opposes the decrease: the near end becomes a south pole, which attracts the receding north pole. Work must be done against this attraction, which supplies the electrical energy.
Specification checklist
- (108) I can describe how the e.m.f. induced by relative motion of a magnet and a coil depends on the speed, the strength of the magnet, the number of turns and the area of the coil.
- (109) I can explain how a changing current in one coil induces an e.m.f. in a linked coil, what increases it, and why a transformer needs an alternating supply.
- (110) I can state Faraday's law and Lenz's law and use \(\varepsilon = -d(N\phi)/dt\), including average e.m.f.s.
- (110) I can draw and interpret flux-linkage–time and e.m.f.–time graphs, using e.m.f. = minus the gradient.
- (110) I can derive and use \(\varepsilon = Blv\) for a straight conductor moving through a uniform field, and find the direction of the induced current.
- (110) I can explain why a coil rotating in a uniform field produces a sinusoidal e.m.f. and where in the rotation it is largest and zero.
- (110) I can explain eddy currents and magnetic braking using Lenz's law, including the energy transfers.